A number has exactly 32 factors out of which 4 are not compo

Question

A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors (which are not composite) is 30. How many such numbers are possible?

A. 2
B. 3
C. 6
D. 8
E. Not possible

navkaran · Accepted Answer

4 are not composite factors.
Product of these 4 = 30
if we start with the first prime numbers 2*3*5 that makes it 30.
So the 4 numbers are 1,2,3,5. And all 4 are not composite

Also number of factors for  2^a * 3^b * 5^c  is (a+1)(b+1)(c+1)=32
and a,b,c cannot be 0
so possible combinations of a,b,c are combinations of 7,1,1 (8*2*2=32) and 3,3,2(4*4*2=32)
Each can have 3 combinations.

So this makes it 6 possible numbers

jonnyhh · Answer

The second sentence tells us that the product of 4 prime factors of the original number is 30. The product of 2, 3 and 5 is itself equal to 30, there is no possibility of a fourth factor here. The answer should be E.

Can you please explain why the OA is C.

mau5 · Answer

Firstly , we should note that 1 is NEITHER a prime nor a composite number.The first composite number is 4.Thus, when the problem states that there are 4 factors that are not composite, these nos are 1,2,3,5. Thus, the given number = 2^a*3^b*5^c. Also, (a+1)*(b+1)*(c+1) = 32. We can break down 32 into 3 integers as : 2*2*8 or 4*4*2
Also, the only possible combinations for a,b,c are : 3,3,1 OR 1,1,7. Thus, each combination has 3 possible orders and we have a total of 6 possibilities.

C.

sanket1991 · Answer

Let's see data,
i) a number with 32 factors, out of which 4 are not composite (Too vague data) 
ii) product of these 4 factors is 30 
lets take (ii) data
30 = 2*3*5*1 (1 is natural number i.e not composite nor prime) 
so we know 3 factors of number are 2,3,5
now 32 = 2*4*4, 2*2*8 
so total numbers possible = 3C1 + 3C1 = 6

chetan2u · Answer

There are two types of number when we consider the number of factors they have:-
1) Composite number: More than 2 factors
2) Prime number: Exactly two factors including 1 and itself
number 1 does not fall in any of the two, as it has only one factor itself.

1 is a factor of all numbers, so the 4 factors that are not composite are 1 and 3 prime numbers =>  n=a^xb^yc^z 
Number of factors =  (x+1)(y+1)(z+1)=32

Product of these 4 factors (which are not composite) is 30  : Irrelevant information, although it tells us that the number is  2^x3^y5^z .

Now we know that the prime numbers are 3: a, b and c.
cases of 32 as product of three numbers greater than 1.

1)  32=2*2*8=(1+1)(1+1)(7+1) 
so x, y and z can be 1, 1 or 7. =  \frac{3!}{2!}=3 
 n=a^1b^1c^7  or  n=a^1b^7c^1  or  n=a^7b^1c^1

2)  32=2*4*4=(1+1)(3+1)(3+1) 
so x, y and z can be 1, 3 or 3. =  \frac{3!}{2!}=3 
 n=a^1b^3c^3  or  n=a^3b^3c^1  or  n=a^3b^1c^3

Total = 3+3=6

C

Kinshook · Answer

Given: A number has exactly 32 factors out of which 4 are not composite. Product of these 4 factors (which are not composite) is 30.

Asked: How many such numbers are possible?

32 = 2ˆ5 = 2*2*2*4
30 = 1*2*3*5

There are 3 prime factors 2, 3, & 5. No other prime factor is possible.

Therefore, the number is of the form  2ˆa*3ˆb*5ˆc 
The number of factors = (a+1)(b+1)(c+1) = 32 = 2*2*8 = 2*4*4

(a,b,c) = {(1,1,7),(1,7,1),(7,1,1),(1,3,3),(3,1,3),(3,1,1)} : Total 6 possibilities

IMO C

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82812 
___
Harsha

pappal · Answer

four non composite factors that result in 30 are 1,2,3,5
to find out the total no of factors of the given no it must be in the powers of 2,3,5
since total no of factors of any prime factorization =(a+1)*(b+1)*(c+1)=32 given
so a,b,c can be either of the form 1,1,7 or 7,1,1 or 1,7,1 OR 1,3,3 or 3,1,3 or 3,3,1
total of six different numbers possible.
OR alternately 1,1,7 and 1,3,3 can be distributed among the powers of 2,3,5 in 3!/!2=3 ways each
i.e. 3+3=6 different permutations are possible resulting in 06 different numbers.
so C

A number has exactly 32 factors out of which 4 are not compo

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